CBSE · NCERT · Class 11 Maths · Chapter 6

NCERT Solutions: Class 11 Maths Chapter 6 - Permutations and Combinations

42 textbook Q&A42 verifiedFree Content

Chapter-wise NCERT intext questions and exercise answers for Permutations and Combinations, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
Sections in this chapter
Exercise 6.1 6Exercise 6.2 5Exercise 6.3 11Exercise 6.4 9Miscellaneous Exercise on Chapter 6 11
Your Progress - Chapter 60% complete
1Exercise 6.16 questions
Q.1How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that (i) repetition of the digits is allowed? (ii) repetition of the digits is not allowed?v
Solution

(i) With repetition allowed, each of the three places can be filled in 5 ways, so the number is $5^3=125$. (ii) Without repetition, the places can be filled in $5,4,3$ ways, so the number is $5\times4\times3=60$.

Answer:

(i) $125$ (ii) $60$.

✓ Verified and validated
Q.2How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?v
Solution

The units digit must be even, so it can be $2,4$ or $6$: 3 choices. Since repetition is allowed, the hundreds and tens places each have 6 choices. Total numbers $=6\times6\times3=108$.

Answer:

$108$.

✓ Verified and validated
Q.3How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?v
Solution

The four positions can be filled in $10,9,8,7$ ways. Hence the number of codes is ${}^{10}P_4=10\times9\times8\times7=5040$.

Answer:

$5040$.

✓ Verified and validated
Q.4How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?v
Solution

The first two digits are fixed as 6 and 7. The remaining three places must be filled from the remaining 8 digits without repetition, so the number is ${}^{8}P_3=8\times7\times6=336$.

Answer:

$336$.

✓ Verified and validated
Q.5A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?v
Solution

Each toss has 2 possible outcomes. For 3 tosses, the number of possible outcomes is $2^3=8$.

Answer:

$8$.

✓ Verified and validated
Q.6Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?v
Solution

Order matters because one flag is below the other. The upper flag can be chosen in 5 ways and the lower flag in 4 ways, so the number of signals is $5\times4=20$.

Answer:

$20$.

✓ Verified and validated
2Exercise 6.25 questions
Q.1Evaluate (i) 8 ! (ii) 4 ! – 3 !v
Solution

$8!=1\times2\times3\times4\times5\times6\times7\times8=40320$. Also $4!-3!=24-6=18$.

Answer:

(i) $8!=40320$ (ii) $4!-3!=18$.

✓ Verified and validated
Q.2Is 3 ! + 4 ! = 7 ! ?v
Solution

$3!=6$ and $4!=24$, so $3!+4!=30$. This is not equal to $7!=5040$.

Answer:

No. $3!+4!=30$, while $7!=5040$.

✓ Verified and validated
Q.3Compute $\dfrac{8!}{6!2!}$v
Solution

$\dfrac{8!}{6!2!}=\dfrac{8\times7\times6!}{6!\times2}=28$.

Answer:

$28$.

✓ Verified and validated
Q.4If $\dfrac{1}{6!}+\dfrac{1}{7!}=\dfrac{x}{8!}$, find xv
Solution

$\dfrac{1}{6!}+\dfrac{1}{7!}=\dfrac{7+1}{7!}=\dfrac{8}{7!}=\dfrac{64}{8!}$. Hence $x=64$.

Answer:

$x=64$.

✓ Verified and validated
Q.5Evaluate $\dfrac{n!}{(n-r)!}$, when (i) n = 6, r = 2 (ii) n = 9, r = 5.v
Solution

(i) $\dfrac{6!}{(6-2)!}=\dfrac{6!}{4!}=6\times5=30$. (ii) $\dfrac{9!}{(9-5)!}=\dfrac{9!}{4!}=9\times8\times7\times6\times5=15120$.

Answer:

(i) $30$ (ii) $15120$.

✓ Verified and validated
3Exercise 6.311 questions
Q.1How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?v
Solution

Fill the three positions from 9 digits without repetition: ${}^{9}P_3=9\times8\times7=504$.

Answer:

$504$.

✓ Verified and validated
Q.2How many 4-digit numbers are there with no digit repeated?v
Solution

The thousands place has 9 choices, from 1 to 9. The remaining three places have 9, 8 and 7 choices from the remaining digits, including 0 where allowed. Total $=9\times9\times8\times7=4536$.

Answer:

$4536$.

✓ Verified and validated
Q.3How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?v
Solution

The units digit must be one of $2,4,6$: 3 choices. After that, the hundreds and tens places have 5 and 4 choices. Total $=3\times5\times4=60$.

Answer:

$60$.

✓ Verified and validated
Q.4Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?v
Solution

Total 4-digit numbers are ${}^{5}P_4=5\times4\times3\times2=120$. For even numbers, the units digit is 2 or 4: 2 choices. The remaining three positions can be filled in ${}^{4}P_3=24$ ways. Thus even numbers $=2\times24=48$.

Answer:

Total $=120$; even numbers $=48$.

✓ Verified and validated
Q.5From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person can not hold more than one position?v
Solution

The chairman can be chosen in 8 ways and the vice chairman in 7 ways. Since the posts are different, total ways $=8\times7=56$.

Answer:

$56$.

✓ Verified and validated
Q.6Find n if ${}^{n-1}P_3 : {}^nP_4 = 1 : 9$.v
Solution

$\dfrac{{}^{n-1}P_3}{{}^nP_4}=\dfrac{(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)}=\dfrac1n$. Since this ratio is $\dfrac19$, $n=9$.

Answer:

$n=9$.

✓ Verified and validated
Q.7Find r if (i) ${}^{5}P_r=2{}^{6}P_{r-1}$ (ii) ${}^{5}P_r={}^{6}P_{r-1}$.v
Solution

(i) $\dfrac{5!}{(5-r)!}=2\dfrac{6!}{(7-r)!}$ gives $(7-r)(6-r)=12$, so $r=3$ is the valid value. (ii) $\dfrac{5!}{(5-r)!}=\dfrac{6!}{(7-r)!}$ gives $(7-r)(6-r)=6$, so $r=4$ is the valid value.

Answer:

(i) $r=3$ (ii) $r=4$.

✓ Verified and validated
Q.8How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?v
Solution

The word EQUATION has 8 distinct letters. The number of arrangements is $8!=40320$.

Answer:

$40320$.

✓ Verified and validated
Q.9How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if. (i) 4 letters are used at a time, (ii) all letters are used at a time, (iii) all letters are used but first letter is a vowel?v
Solution

MONDAY has 6 distinct letters. (i) Four-letter arrangements: ${}^{6}P_4=360$. (ii) All letters: $6!=720$. (iii) The first letter can be one of the two vowels A or O, and the remaining 5 letters can be arranged in $5!$ ways, giving $2\times5!=240$.

Answer:

(i) $360$ (ii) $720$ (iii) $240$.

✓ Verified and validated
Q.10In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?v
Solution

MISSISSIPPI has 11 letters: I appears 4 times, S appears 4 times and P appears 2 times. Total distinct permutations $=\dfrac{11!}{4!4!2!}=34650$. If the four I's come together, treat them as one block; then there are 8 objects with S repeated 4 times and P repeated 2 times, giving $\dfrac{8!}{4!2!}=840$. Therefore, not together $=34650-840=33810$.

Answer:

$33810$.

✓ Verified and validated
Q.11In how many ways can the letters of the word PERMUTATIONS be arranged if the (i) words start with P and end with S, (ii) vowels are all together, (iii) there are always 4 letters between P and S?v
Solution

PERMUTATIONS has 12 letters with T repeated twice. (i) Fix P first and S last; arrange the remaining 10 letters with two T's: $\dfrac{10!}{2!}=1814400$. (ii) Treat the 5 vowels E, U, A, I, O as one block. Then arrange this block and 7 consonants, with T repeated twice: $\dfrac{8!}{2!}\times5!=2419200$. (iii) P and S must occupy positions differing by 5. There are 7 such position-pairs and 2 orders for P and S, so 14 choices. The remaining 10 letters can be arranged in $\dfrac{10!}{2!}$ ways. Total $=14\times\dfrac{10!}{2!}=25401600$.

Answer:

(i) $1814400$ (ii) $2419200$ (iii) $25401600$.

✓ Verified and validated
4Exercise 6.49 questions
Q.1If nC8 = nC2, find nC2.v
Solution

Since ${}^{n}C_8={}^{n}C_2$, either $8=2$ or $8=n-2$. Hence $n=10$. Therefore ${}^{n}C_2={}^{10}C_2=45$.

Answer:

${}^{n}C_2=45$.

✓ Verified and validated
Q.2Determine n if (i) ${}^{2n}C_3 : {}^{n}C_3 = 12 : 1$ (ii) ${}^{2n}C_3 : {}^{n}C_3 = 11 : 1$v
Solution

$\dfrac{{}^{2n}C_3}{{}^{n}C_3}=\dfrac{2n(2n-1)(2n-2)}{n(n-1)(n-2)}=\dfrac{4(2n-1)}{n-2}$. (i) Set this equal to 12: $4(2n-1)=12(n-2)$, so $n=5$. (ii) Set it equal to 11: $4(2n-1)=11(n-2)$, so $n=6$.

Answer:

(i) $n=5$ (ii) $n=6$.

✓ Verified and validated
Q.3How many chords can be drawn through 21 points on a circle?v
Solution

A chord is determined by choosing any 2 of the 21 points. Hence the number of chords is ${}^{21}C_2=210$.

Answer:

$210$.

✓ Verified and validated
Q.4In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?v
Solution

Choose 3 boys from 5 and 3 girls from 4: ${}^{5}C_3\times{}^{4}C_3=10\times4=40$.

Answer:

$40$.

✓ Verified and validated
Q.5Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.v
Solution

Select 3 red from 6, 3 white from 5 and 3 blue from 5. The number of ways is ${}^{6}C_3\times{}^{5}C_3\times{}^{5}C_3=20\times10\times10=2000$.

Answer:

$2000$.

✓ Verified and validated
Q.6Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.v
Solution

Choose 1 ace from 4 and 4 non-aces from the remaining 48 cards. Number of hands $={}^{4}C_1\times{}^{48}C_4=4\times194580=778320$.

Answer:

$778320$.

✓ Verified and validated
Q.7In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?v
Solution

Choose exactly 4 bowlers from 5 and the remaining 7 players from the 12 non-bowlers. Number of teams $={}^{5}C_4\times{}^{12}C_7=5\times792=3960$.

Answer:

$3960$.

✓ Verified and validated
Q.8A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.v
Solution

Choose 2 black balls from 5 and 3 red balls from 6: ${}^{5}C_2\times{}^{6}C_3=10\times20=200$.

Answer:

$200$.

✓ Verified and validated
Q.9In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?v
Solution

The 2 compulsory courses are fixed. The student must choose 3 more courses from the remaining 7, which can be done in ${}^{7}C_3=35$ ways.

Answer:

$35$.

✓ Verified and validated
5Miscellaneous Exercise on Chapter 611 questions
Q.1How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER ?v
Solution

DAUGHTER has 3 vowels A, U, E and 5 consonants D, G, H, T, R. Choose 2 vowels and 3 consonants, then arrange the 5 selected letters: ${}^{3}C_2\times{}^{5}C_3\times5!=3\times10\times120=3600$.

Answer:

$3600$.

✓ Verified and validated
Q.2How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?v
Solution

EQUATION has 5 vowels and 3 consonants. Treat the vowels as one block and the consonants as another block; the two blocks can be arranged in $2!$ ways. The vowels can be arranged in $5!$ ways and consonants in $3!$ ways. Total $=2!\times5!\times3!=1440$.

Answer:

$1440$.

✓ Verified and validated
Q.3A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of: (i) exactly 3 girls ? (ii) atleast 3 girls ? (iii) atmost 3 girls ?v
Solution

(i) Exactly 3 girls: ${}^{4}C_3\times{}^{9}C_4=4\times126=504$. (ii) At least 3 girls means 3 girls or 4 girls: $504+{}^{4}C_4{}^{9}C_3=504+84=588$. (iii) At most 3 girls means 0,1,2 or 3 girls: ${}^{9}C_7+{}^{4}C_1{}^{9}C_6+{}^{4}C_2{}^{9}C_5+{}^{4}C_3{}^{9}C_4=36+336+756+504=1632$.

Answer:

(i) $504$ (ii) $588$ (iii) $1632$.

✓ Verified and validated
Q.4If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E ?v
Solution

In dictionary order, words before the first word starting with E must start with A. Fix A first. The remaining 10 letters contain A once, I twice and N twice. Hence the number of such words is $\dfrac{10!}{2!2!}=907200$.

Answer:

$907200$.

✓ Verified and validated
Q.5How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated ?v
Solution

A number divisible by 10 must end in 0. The remaining five positions are filled by 1, 3, 5, 7 and 9 in any order, giving $5!=120$ numbers.

Answer:

$120$.

✓ Verified and validated
Q.6The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet ?v
Solution

Choose 2 vowels from 5 and 2 consonants from 21, then arrange the 4 selected letters: ${}^{5}C_2\times{}^{21}C_2\times4!=10\times210\times24=50400$.

Answer:

$50400$.

✓ Verified and validated
Q.7In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions ?v
Solution

The possible splits are $(3,5)$, $(4,4)$ and $(5,3)$ from Part I and Part II. Number of choices $={}^{5}C_3{}^{7}C_5+{}^{5}C_4{}^{7}C_4+{}^{5}C_5{}^{7}C_3=10\times21+5\times35+1\times35=420$.

Answer:

$420$.

✓ Verified and validated
Q.8Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.v
Solution

Choose exactly 1 king from 4 and the remaining 4 cards from the 48 non-kings. Number of selections $={}^{4}C_1\times{}^{48}C_4=4\times194580=778320$.

Answer:

$778320$.

✓ Verified and validated
Q.9It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible ?v
Solution

In 9 places, the even places are 2, 4, 6 and 8. The 4 women can occupy these in $4!$ ways, and the 5 men can occupy the remaining 5 places in $5!$ ways. Total $=4!\times5!=2880$.

Answer:

$2880$.

✓ Verified and validated
Q.10From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen ?v
Solution

If the 3 students all join, choose the remaining 7 from the other 22: ${}^{22}C_7=170544$. If none of them joins, choose all 10 from the other 22: ${}^{22}C_{10}=646646$. Total $=170544+646646=817190$.

Answer:

$817190$.

✓ Verified and validated
Q.11In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together ?v
Solution

ASSASSINATION has 13 letters: A appears 3 times, S appears 4 times, I appears 2 times and N appears 2 times. Keep all S's as one block. Then there are 10 objects: the S-block, A,A,A, I,I, N,N, T, O. The number of arrangements is $\dfrac{10!}{3!2!2!}=151200$.

Answer:

$151200$.

✓ Verified and validated
Back: Ch.5
Next: Ch.7